Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). Now, I assume the oscillations are small and so the small angle approximation holds. I wish to find the period of oscillation and I analyze the motion in two ways, first using conservation of energy and secondly using dynamics. However, I receive two inconsistent answers. One or both of the solutions must be wrong, but I cannot figure out which one and more importantly, I cannot figure out why.

Method 1: We write the energy conservation equation for the ball

$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = Constant$

from the center of mass, we take the height as $h = R-(R-r)cos\theta$ where $\theta$ is the angle from the vertical. Applying the no slip condition $v = r\omega$ and taking the moment of inertia for a solid sphere $I = \frac{2}{5}mr^2$ we can write the energy equation as

$mg(R-(R-r)cos\theta) + \frac{7}{10}mr^2\omega^2 = Constant$

Differentiating with respect to time:

$mg(R-r)sin\theta\cdot\omega + \frac{7}{5}mr^2\omega\cdot\alpha = 0$

taking the small angle approximation $sin\theta = \theta$ we get

$g(R-r)\theta + \frac{7}{5}r^2\alpha=0$

$-\frac{5g(R-r)}{7r^2}\theta = \alpha$

from which we can get $T = 2\pi\sqrt{\frac{7r^2}{5g(R-r)}}$

Method 2: The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write

$\tau = I\alpha = fr$

again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,

$\frac{2}{5}ma = f$

The net force acting on the system is the tangential component of gravity and the force of friction, so

$F = ma = mgsin\theta - f$

$\frac{7}{5}a = gsin\theta$

taking the small angle approximation and converting $a$ to $\alpha$ we get

$\alpha = \frac{5g}{7r}\theta$

and a corresponding period of $T = 2\pi\sqrt{\frac{7r}{5g}}$

Now the solutions are very different and I would appreciate it if someone would point out where I went wrong.

  • $\begingroup$ There are a lot of things wrong with this. The most glaring is that you take energy to be constant in the first approach, but you introduce friction into the second, which is an outside force that will reduce energy over time. So the two approaches are necessarily going to give different results because you're making different assumptions. $\endgroup$
    – Mitchell
    Jun 18, 2011 at 5:25
  • 5
    $\begingroup$ @Mitchell, no, that is not a problem. The friction force does no work. I think Yuqing explained this pretty clearly. $\endgroup$ Jun 18, 2011 at 5:33
  • $\begingroup$ Closely related: physics.stackexchange.com/q/10798 $\endgroup$
    – qftme
    Jun 18, 2011 at 12:23

4 Answers 4


Your first derivation, using energy, uses two different meanings for the same symbol $\omega$. In one place, you interpret it as

$$\omega = \dot{\theta}$$

the time derivative of the angle of the line from the center of the ball to the center of the bowl with the vertical.

In another place, you interpret $\omega$ as the time derivative of the unnamed angle through which the ball itself has rotated.

These two angles are related to each other by the $r/(R-r)$ factor by which you are off.

  • 1
    $\begingroup$ Hi, I had the exact same confusion as the OP but your answer really cleared things up for me. Only thing I still can't work out is why the two angles are related to each other by the $r(R-r)$ factor. Some help? $\endgroup$
    – GeeJay
    Nov 24, 2016 at 12:46
  • $\begingroup$ @GeeJay The center of the ball is moving with a velocity $(R-r) \dot{\theta}$. And the angular velocity of the ball of radius $r$ is $\omega$. Then, the net tangetial velocity of the point of contact is $(R-r) \dot{\theta} -r \omega$, which must be zero, because there is no slipping. $\endgroup$ Jul 1, 2021 at 17:48

The answer you got in first method was wrong.

The answer in second is nearly reached, we have to take $R-r$ instead of $r$ because you have to take $r$ from centre of the sphere. The correct answer was $2\pi\sqrt{7(R-r)/5g}$

  • 1
    $\begingroup$ Hi and welcome to the Physics SE! The equations become much more readable and searchable with mathjax. It'd be great if you could use it in your next posts. $\endgroup$
    – stafusa
    Dec 14, 2017 at 14:39

A problem I see in your first solution is that $v$ is the velocity of the center of mass. You assume $v=r\omega$ but that is the velocity of the bottom of the ball. The center of mass is moving parallel to this in a circular motion, the radius of which is $R-r$, $v=((R-r)/R)r\omega$.

  • $\begingroup$ $v$ of the center of mass is also $(R - r) \dot{\theta}$. Your relation implies $(R - r) \dot{\theta} = \frac{R-r}{R} r \omega$, or $\dot{\theta} = \frac{r}{R} \omega$, which is wrong (consider the case where $r$ and $R$ are almost equal. There $\dot{\theta} \neq 0$, but $\omega \approx 0$). The correct relation is $(R-r) \dot{\theta} = r \omega$. $\endgroup$ Jul 1, 2021 at 17:57

Both answers are incorrect. The correct answer taking into account both angles as mentioned by others, has R-r in the numerator. The limiting case, in which the ball gets really small, produces a period that cannot go to zero, and this consideration alone can be used to eliminate both of your answers.


Not the answer you're looking for? Browse other questions tagged or ask your own question.